Question

The compressibility factor for $N_2$ at $223K$ and $81.06MPa$ is $1.95$, and at $373K$ and $20.265MPa$ it is $1.10$. A certain mass of $N_2$ occupies a volume of $1.0d{m}^3$ at $223K$ and $81.06MPa$. What is the volume occupied by the same quantity of $N_2$ at $373K$ and $20.265MPa$?

• A. $3.774d{m}^3$
• B. $2.77d{m}^3$
• C. $5.07d{m}^3$
• D. $9.30d{m}^3$

Answer 1

The correct option is A $3.774d{m}^3$

For $T=223K,P=81.06MPa,Z=1.95$, and $V=1.0d{m}^3={10}^{3}c{m}^3$, we have
$n=\frac{PV}{ZRT}=\frac{81.06\times {10}^3}{1.95\times 8.314\times 223}=22.42mol$
Now, at $T=373K,P=20.265MPa,Z=1.10$, the volume occupied will be
$V=\frac{ZnRT}{P}=\frac{1.10\times 22.42\times 8.314\times 373}{20.265}=3774.0c{m}^3$
$\therefore V=3.774d{m}^3$.